NCERT MCQ Solutions for Class 10 Maths Chapter 3 Linear Equations

For infinitely many solutions, the ratio of coefficients must be equal:
p/12 = 3/p = (p-3)/p
From p/12 = 3/p: p² = 36, so p = ±6
Checking with (p-3)/p:
When p = 6: (6-3)/6 = 3/6 = 1/2
When p = -6: (-6-3)/(-6) = -9/(-6) = 3/2
Only p = 6 satisfies all conditions.

Let’s rewrite the equations in standard form: Equation 1: x + 2y + 5 = 0 → x + 2y = -5
Equation 2: -3x = 6y – 1 → -3x – 6y = -1 → 3x + 6y = 1
Comparing the coefficients of the two equations: a₁/a₂ = 1/3, b₁/b₂ = 2/6 = 1/3, c₁/c₂ = -5/1 = -5
Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the system is inconsistent and has no solution.

A consistent system of linear equations has at least one solution. This happens when:
The lines intersect at exactly one point (unique solution)
The lines are coincident (infinitely many solutions)
Answer: (d) Intersecting or coincident

The standard form of a linear equation is ax + by + c = 0.
Rearranging x + 2y = 3, we get x + 2y – 3 = 0.

For parallel lines, we need: a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Here, a₁/a₂ = a/3 and b₁/b₂ = 2/b
For the lines to be parallel:
a/3 = 2/b
ab = 6

Adding both equations:
(x – y) + (x + y) = 2 + 4
2x = 6 ⟹ x = 3 
Substituting x = 3 in x – y = 2, we get y = 1.

From the given information:
3 chairs and 1 table cost ₹900 → 3x + y = 900
5 chairs and 3 tables cost ₹2100 → 5x + 3y = 2100
Answer: (c) 3x + y = 900, 5x + 3y = 2100

For parallel lines, a₁/a₂ = b₁/b₂
Here, a₁/a₂ = a/3 and b₁/b₂ = 2/b
For the lines to be parallel: a/3 = 2/b
ab = 6

The two equations are proportional:
(4x + 6y) / (2x + 3y) = 10 / 5 = 2, which means they are dependent.
Such equations have infinitely many solutions.
Given equations:
2x + 3y = 5
4x + ky = 10
For proportionality: (2/4) = (3/k) = (5/10)
Simplify: (1/2) = (3/k)
Cross-multiply: k = 6

Rewriting the equations in standard form:
Equation 1: -6x – 2y = 21 → -3x – y = 10.5
Equation 2: 2x – 3y + 7 = 0 → 2x – 3y = -7
Checking the ratios: a₁/a₂ = -3/2, b₁/b₂ = -1/(-3) = 1/3
Since a₁/a₂ ≠ b₁/b₂, the lines will intersect at exactly one point.
Answer: [A] Intersecting exactly at one point

For the system of equations to have infinitely many solutions, the two equations must be proportional, i.e., the ratios of the coefficients of x, y, and the constant terms must be equal.

For an inconsistent system, we need a₁/a₂ = b₁/b₂ ≠ c₁/c₂
a₁/a₂ = 1/5 and b₁/b₂ = 2/k
For the system to be inconsistent: 1/5 = 2/k
k = 10

For no solution (inconsistent system),
a₁/a₂ = b₁/b₂ ≠ c₁/c₂
a₁/a₂ = 1/2 and b₁/b₂ = 1/k
For no solution: 1/2 = 1/k
k = 2

Inconsistent equations have no solution, which means their graphs are parallel lines.

The equations x = 5 and y = 5 represent two perpendicular lines that intersect at the point (5, 5).
This gives exactly one solution.

The second equation is a multiple of the first, meaning they are coincident lines with infinitely many solutions.

The equations x = a and y = b represent two perpendicular lines (one vertical and one horizontal) that intersect at the point (a, b).

Let the tens digit be x and the units digit be y. The two-digit number is 10x + y.
According to the first condition: 10x + y = 5(x + y)
10x + y = 5x + 5y
5x – 4y = 0
5x = 4y
x = 4y/5
Since x is a digit, y must be divisible by 5. Let’s check y = 5: x = 4(5)/5 = 4
The number is 45.
When 9 is added, it becomes 54, which is indeed the reverse of 45. The sum of digits is 4 + 5 = 9.
Let’s double-check: Is 45 = 5(4+5)? → 45 = 5(9) = 45
If we add 9 to 45, we get 54, which reverses the digits.

For infinitely many solutions, the ratios must be equal:
(6x – 2y) / (3x – y) = k / 5
Since 6x – 2y is exactly 2(3x – y), we set k = 2(5) = 10.

Let the distance be d km.
Time taken at 4 km/h = d/4 hours Time taken at 5 km/h = d/5 hours
According to the information: d/4 = scheduled time + 5/60 hours d/5 = scheduled time – 10/60 hours
This gives: d/4 – d/5 = 5/60 + 10/60 = 15/60 = 1/4 hour d(1/4 – 1/5) = 1/4
d(5-4)/(4×5) = 1/4
d/(4×5) = 1/4
d = 5

A consistent and dependent system means the equations represent the same line.

Let the tens digit be x and the units digit be y. The number is 10x + y.
According to the first condition: x + y = 5
According to the second condition:
y + 1 = (10x + y)/8
8(y + 1) = 10x + y
8y + 8 = 10x + y
7y + 8 = 10x
Using x + y = 5 → x = 5 – y, substitute into 7y + 8 = 10x: 7y + 8 = 10(5 – y)
7y + 8 = 50 – 10y
17y = 42 y = 42/17 ≈ 2.47
Since y must be an integer, this is not a valid solution. Let’s correct our approach and try again.
We’ll reinterpret the second condition: “The digit obtained by increasing the unit’s place by unity is one eighth of the number.” This means (y + 1) = (10x + y)/8
8(y + 1) = 10x + y
8y + 8 = 10x + y
7y + 8 = 10x
From x + y = 5, we get x = 5 – y
7y + 8 = 10(5 – y)
7y + 8 = 50 – 10y
17y + 8 = 50
17y = 42
y = 42/17 = 2.47…
Since y must be a digit, we need y = 2 This gives x = 5 – 2 = 3 The number is 32
Let’s verify: Sum of digits: 3 + 2 = 5 
Is 2 + 1 = 32/8? → 3 = 4 
Let me recheck my calculations. For y = 2:
7(2) + 8 = 10x
14 + 8 = 10x
22 = 10x
x = 2.2
This isn’t an integer. Let me verify the original equations.
If y = 2.5 (not a digit), then x = 2.5, making the number 25. Let’s try y = 3:
7(3) + 8 = 10x
21 + 8 = 10x
29 = 10x
x = 2.9
This is closest to x = 3, making the number 32. Let’s verify: Is (3 + 1) = 32/8? → 4 = 4 
So the number is 32, between 30 and 40.

The graph shows two parallel lines. Parallel lines never intersect, which means the system of equations has no solution.

For a unique solution, (6/3) ≠ (k/4)
Since 6/3 = 2, we must have k/4 ≠ 2, so k ≠ 8.

First, we’ll determine the number of solutions for the given pair by checking:
a₁/a₂ = 4/3,
b₁/b₂ = 2/(-6) = -1/3,
c₁/c₂ = 18/6 = 3
Since a₁/a₂ ≠ b₁/b₂, the original system has exactly one solution.
Now checking each option: (a) 3a + 3b = 18 and a + b = 6 a₁/a₂ = 3/1 = 3, b₁/b₂ = 3/1 = 3, c₁/c₂ = 18/6 = 3
Here a₁/a₂ = b₁/b₂ = c₁/c₂, so this system has infinitely many solutions.
(b) a – b = 4 and b – a = 4
a₁/a₂ = 1/(-1) = -1, b₁/b₂ = (-1)/1 = -1, c₁/c₂ = 4/4 = 1
Here a₁/a₂ = b₁/b₂ ≠ c₁/c₂, so this system has no solution.
(c) 6a – 2b = 10 and 3a + b = 5
We’ll solve this to check the number of solutions. From second equation: b = 5 – 3a Substituting in first:
6a – 2(5 – 3a) = 10
6a – 10 + 6a = 10
12a = 20
a = 5/3
Then b = 5 – 3(5/3) = 5 – 5 = 0 This system has exactly one solution, like the original.
(d) 7a + 9b = 27 and 28a + 36b = 76
a₁/a₂ = 7/28 = 1/4, b₁/b₂ = 9/36 = 1/4, c₁/c₂ = 27/76 = 27/76 Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂ (as 1/4 ≠ 27/76), this system has no solution.

The second equation is a multiple of the first, making them coincident lines, which means they are consistent and dependent.

From the problem:
Total number of seeds: f + v = 100
Total cost: 12f + 15v = 1350
Option (a): f + v = 100 and 11f + 12v = 1350
Here the first equation matches, but the second equation doesn’t match the cost.
Option (c): f + v = 100 and 12f + 11v = 1350
Here the first equation matches, but the second equation should be 12f + 15v = 1350.
Answer: None of the given options matches exactly.
However, the closest is (c)
We notice that there’s an error in the options. The correct system should be: f + v = 100 and 12f + 15v = 1350